What modifications do I need to make in the following codes for solving the boundary value problem similar to the Blasius equation using Shooting method with R-K 4 numerical analysis
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
The equation is (1+2M*eta)f'''+ 2Mf"+ f*f"- (f')^2- K1*f'= 0 ; f' is df/d(eta) 'eta' is a similarity variable
where M- is a curvature parameter, M = 1
k1- is another parameter, k1= 0.1
boundary conditions are: eta=0, f(0)=0, f'(0)=1, f'(inf)=0, inf= 10
The following is the code for Blasius equation
% Fourth Runge Kutta for solving Blasius Equation
%-----------------------------------------------
%Blasius Equation : 2f"+ff'=0
%-----------------------------------------------
%Boundary Condition
%1. f'(0) = 0
%2. f'(inf) = 0
%3. f(0) = 0
%from shooting method
%4. f"(0) = 0.332
%-----------------------------------------------
%find: f,f" and f'''
%-----------------------------------------------
%Given
%eta = x
%f = y0,f'=y1,f"=y2,f'''= -(1/2)*y0*y2
%-----------------------------------------------
func1 = inline('y1','x','y0','y1','y2');
%-----------------------------------------------
func2 = inline('y2','x','y0','y1','y2');
%-----------------------------------------------
func3 = inline('-0.5*y0*y2','x','y0','y1','y2');
%-----------------------------------------------
%input: x = 0 , y0 = 0 , y1= 0
% y2 = 0.332 , total = 7 and h = 0.1
%-----------------------------------------------
x = input('\n Enter the value of x : ');
y0 = input( 'Enter the value of y0 : ');
y1 = input( 'Enter the value of y1 : ');
y2 = input( 'Enter the value of y2 : ');
total = input('Enter the value of total : ');
h = input( 'Enter the value of h : ');
fprintf('\n Solution with step size =%5.3f is:',h);
fprintf('\n x y0 y1 y2');
fprintf('\n%16.3f%16.3f%16.3f%16.3f',x,y0,y1,y2);
for i = 1:(total/h)
ak1y0 = func1(x,y0,y1,y2);
ak1y1 = func2(x,y0,y1,y2);
ak1y2 = func3(x,y0,y1,y2);
xx = x + h/2.;
yy0 = y0 + h*ak1y0/2.;
yy1 = y1 + h*ak1y1/2.;
yy2 = y2 + h*ak1y2/2.;
ak2y0 = func1(xx,yy0,yy1,yy2);
ak2y1 = func2(xx,yy0,yy1,yy2);
ak2y2 = func3(xx,yy0,yy1,yy2);
yy0 = y0 + h*ak2y0/2.;
yy1 = y1 + h*ak2y1/2.;
yy2 = y2 + h*ak2y2/2.;
ak3y0 = func1(xx,yy0,yy1,yy2);
ak3y1 = func2(xx,yy0,yy1,yy2);
ak3y2 = func3(xx,yy0,yy1,yy2);
all_x(i) = x;
all_y0(i) = y0;
all_y1(i) = y1;
all_y2(i) = y2;
xx = x + h;
yy0 = y0 + h*ak3y0;
yy1 = y1 + h*ak3y1;
yy2 = y2 + h*ak3y2;
ak4y0 = func1(xx,yy0,yy1,yy2);
ak4y1 = func2(xx,yy0,yy1,yy2);
ak4y2 = func3(xx,yy0,yy1,yy2);
y0 = y0 + (ak1y0 + 2.*ak2y0 + 2.*ak3y0 + ak4y0)*h/6.;
y1 = y1 + (ak1y1 + 2.*ak2y1 + 2.*ak3y1 + ak4y1)*h/6.;
y2 = y2 + (ak1y2 + 2.*ak2y2 + 2.*ak3y2 + ak4y2)*h/6.;
x = x + h;
fprintf('\n%16.3f%16.3f%16.3f%16.3f',x,y0,y1,y2);
end
plot(all_x, all_y0, 'k-', all_x, all_y1, 'b-', all_x, all_y2, 'g-')
0 Kommentare
Akzeptierte Antwort
Torsten
am 8 Nov. 2017
Use MATLAB's "bvp4c" instead of a code that you don't understand.
Best wishes
Torsten.
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Fluid Dynamics finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!